Now we are ready to introduce the most important concept of the projection of a vector onto an axis. It is constantly used in solving physical problems.
7.5.1 What is the projection of a vector onto an axis?
Let the vector ~a and the X-axis be given. It is assumed that the X-axis has a scale that allows you to measure the lengths of the segments and assign them the dimension of the vector ~a.
From the beginning and end of the vector ~a we drop perpendiculars to the X axis; let A and B be the bases of these perpendiculars (Fig. 7.26). Denote the length of the segment AB by jABj.
Rice. 7.26. Projection of a vector onto an axis
Definition. The projection ax of the vector ~a onto the X axis is equal to the length of the segment AB, taken with a plus sign if the angle " between the vector ~a and the X axis is acute, and taken with a minus sign, respectively, if " is obtuse (or unfolded). If the angle is right, then ax = 0.
In short, we have the following formula:
Figure 7.27 illustrates all of these possibilities.
Here, as usual, a = j~aj is the modulus of the vector ~a.
Indeed, if "< 90 , то формула (7.10 ) даёт длину левого красного отрезка на рис.7.27 .
If "\u003e 90, then, moving in the middle part of Fig. 7.27 to the angle adjacent to the angle ", we see that formula (7.10) gives the length of the middle red segment with a minus sign (due to the negativity of the cosine), which is exactly what we need need.
Finally, if " = 90 , then formula (7.10 ) gives ax = 0, since the cosine right angle equals zero. This is exactly how it should be (right side of the figure).
Suppose now that the x-axis is given an additional origin, so that it is the usual coordinate axis. Then we have one more formula for the projection ax , which also contains all three cases of Figure 7.27 in a ¾archived¿ form.
Corollary 2. Let x1 and x2 be the coordinates of the beginning and end of the vector ~a, respectively. Then the projection ax is calculated by the formula:
ax = x2 x1 : |
Indeed, let's look at Fig. 7.28. This is a case of positive projection. It is obvious from the figure that the difference x2 x1 is equal to the length of the red segment, and this length in this case is precisely the projection ax.
Rice. 7.28. Projection of a vector onto an axis. To Corollary 2
What will happen in the remaining two cases (ax< 0 и ax = 0)? Убедитесь, пожалуйста, самостоятельно, что формула (7.11 ) и для них остаётся справедливой.
7.5.2 Vector to Axis Projection Properties
The operation of projecting a vector onto an axis agrees remarkably well with the operations of vector addition and scalar-vector multiplication. Namely, whatever the x-axis, the following two design properties hold.
1. The projection of the vector ~a + b onto the X axis is ax + bx .
Brief verbal formulation: the projection of the sum of vectors is equal to the sum of their projections. This is true for the sum of any number of vectors, not just two.
Rice. 7.29. ~c = ~a + b) cx = ax | |||||||||
First of all, we illustrate this statement in the figure. Let's place the beginning of the century-
of the torus b to the end of the vector ~a, and let ~c = ~a + b (Fig. 7.29).
On the this figure it is clearly seen that the projection cx is equal to the sum of the lengths of the red and green segments, that is, just ax + bx.
True, fig. 7.29 is made for the case ax > 0 and bx > 0. To prove our assertion for all at once possible values projections ax and bx , we will carry out the following universal reasoning based on the formula (7.11).
So, let the vectors ~a and b be located arbitrarily. Again compatible start |
||
of the vector b with the end of the vector ~a and denote ~c = ~a + b. Let: |
||
the coordinate of the beginning of the vector ~a and at the same time the beginning of the vector ~c; |
||
the coordinate of the end of the vector ~a and at the same time the beginning of the vector b; |
||
coordinate of the end of the vector b and at the same time the end of the vector ~c. |
||
These designations are also present in Fig. 7.29.
By formula (7.11 ) we have: ax = x2 x1 , bx = x3 x2 , cx = x3 x1 . Now it is easy to see that:
ax + bx = (x2 x1 ) + (x3 x2 ) = x3 x1 = cx :
Our first projection property is thus proved.
2. The projection of the vector ~a on the X axis is a x .
Verbal formulation: the projection of the product of a scalar and a vector is equal to the product of a scalar and the projection of a vector.
Let's start again with an illustration. The left side of Figure 7.30 shows a vector ~a with a positive projection ax .
Rice. 7.30. The projection of the vector ~a is equal to ax
If you multiply the vector ~a by 2, then its length will double, the projection of the vector will also double (preserving the sign) and become equal to 2ax .
If we multiply the vector ~a by 2, then its length will again double, but the direction will be reversed. The projection will change sign and become equal to 2ax .
Thus, the essence of the second property is clear, and now we can give a rigorous proof.
So let ~ . We go to prove that x x . b = ~a b = a
Let us use formula (7.10) for this. We have:
ax = a cos "; bx = b cos ;
where is the angle between the vector and the axis, and the angle between the vector ~ and the axis. Except
Moreover, by virtue of the definition of multiplication of a scalar by a vector:
In this way:
bx = j ja cos:
If, then j j ; in this case the vector ~ is co-directed with the vector, and therefore.
> 0 = b~a = "
bx = a cos" = ax:
If, then j j ; in this case, the vector ~ is opposite in the direction of the vector
ru ~a. It is easy to figure out that = " (for example, if " is sharp, that is, adjacent to it is obtuse, and vice versa). We then have:
bx = ()a cos(") = ()a(cos ") = a cos " = ax :
Thus, in all cases, the desired relation is obtained, and thus the second property of the projection is completely proved.
7.5.3 Design operation in physics
The proven properties of the design operation are very important for us. In mechanics, for example, we will use them at every turn.
Thus, the solution of many problems in dynamics begins with writing Newton's second law in vector form. Take, for example, a pendulum of mass m suspended on a thread. For a pendulum, Newton's second law will be:
Having written Newton's second law in vector form, we proceed to its projection on
suitable axes. We take equality (7.12) and project onto the X axis: | |
max = mgx + Tx + fx : |
When passing from vector equality (7.12 ) to scalar equality (7.13 ), both design properties are used! Namely, due to property 1, we have written the projection of the sum of vectors as the sum of their projections; property 2 allows us to write the projections of the vectors m~a and m~g as max and mgx .
Thus, both properties of the projection operation ensure the transition from vector to scalar equalities, and this transition can be performed formally and without thinking: we discard the arrows in the notation of vectors and put projection indices instead. This is exactly what the transition from equation (7.12) to equation (7.13) looks like.
Algebraic vector projection on any axis is equal to the product of the length of the vector and the cosine of the angle between the axis and the vector:Right a b = |b|cos(a,b) or
Where a b is the scalar product of vectors , |a| - modulus of vector a .
Instruction. To find the projection of the vector Пp a b in online mode you must specify the coordinates of the vectors a and b . In this case, the vector can be given in the plane (two coordinates) and in space (three coordinates). The resulting solution is saved in a Word file. If the vectors are given through the coordinates of the points, then you must use this calculator.
Vector projection classification
Types of projections by definition vector projection
Types of projections by coordinate system
Vector projection properties
- The geometric projection of a vector is a vector (it has a direction).
- The algebraic projection of a vector is a number.
Vector projection theorems
Theorem 1. The projection of the sum of vectors on any axis is equal to the projection of the terms of the vectors on the same axis.
Theorem 2. The algebraic projection of a vector onto any axis is equal to the product of the length of the vector and the cosine of the angle between the axis and the vector:
Right a b = |b|cos(a,b)
Types of vector projections
- projection onto the OX axis.
- projection onto the OY axis.
- projection onto a vector.
| Projection onto the OX axis | Projection onto the OY axis | Projection to vector |
If the direction of the vector A'B' coincides with the direction of the OX axis, then the projection of the vector A'B' has a positive sign.  | If the direction of the vector A'B' coincides with the direction of the OY axis, then the projection of the vector A'B' has a positive sign.  | If the direction of the vector A'B' coincides with the direction of the vector NM, then the projection of the vector A'B' has a positive sign.  |
If the direction of the vector is opposite to the direction of the OX axis, then the projection of the vector A'B' has a negative sign.  | If the direction of the vector A'B' is opposite to the direction of the OY axis, then the projection of the vector A'B' has a negative sign.  | If the direction of the vector A'B' is opposite to the direction of the vector NM, then the projection of the vector A'B' has a negative sign.  |
If the vector AB is parallel to the axis OX, then the projection of the vector A'B' is equal to the modulus of the vector AB.   | If the vector AB is parallel to the OY axis, then the projection of the vector A'B' is equal to the modulus of the vector AB.   | If the vector AB is parallel to the vector NM, then the projection of the vector A'B' is equal to the modulus of the vector AB.   |
If the vector AB is perpendicular to the axis OX, then the projection of A'B' is equal to zero (zero-vector).   | If the vector AB is perpendicular to the OY axis, then the projection of A'B' is equal to zero (a null vector).   | If the vector AB is perpendicular to the vector NM, then the projection of A'B' is equal to zero (a null vector).   |
1. Question: Can the projection of a vector have a negative sign. Answer: Yes, vector projections can be negative. In this case, the vector has the opposite direction (see how the OX axis and the AB vector are directed)
2. Question: Can the projection of a vector coincide with the modulus of the vector. Answer: Yes, it can. In this case, the vectors are parallel (or lie on the same line).
3. Question: Can the projection of a vector be equal to zero (zero-vector). Answer: Yes, it can. In this case, the vector is perpendicular to the corresponding axis (vector).
Example 1 . The vector (Fig. 1) forms an angle of 60 o with the OX axis (it is given by the vector a). If OE is a scale unit, then |b|=4, so 
.
Indeed, the length of the vector ( geometric projection b) is equal to 2, and the direction is the same as the direction of the OX axis.
Example 2 . The vector (Fig. 2) forms an angle with the OX axis (with the vector a) (a,b) = 120 o . Length |b| vector b is equal to 4, so pr a b=4 cos120 o = -2. 
Indeed, the length of the vector is equal to 2, and the direction is opposite to the direction of the axis.
projection vector on an axis is called a vector, which is obtained by multiplying the scalar projection of a vector on this axis and the unit vector of this axis. For example, if a x is scalar projection vector a on the x-axis, then a x i- its vector projection on this axis.
Denote vector projection just like the vector itself, but with the index of the axis on which the vector is projected. So, the vector projection of the vector a on the x-axis denote a x ( oily a letter denoting a vector and a subscript of the axis name) or (a non-bold letter denoting a vector, but with an arrow at the top (!) and a subscript of the axis name).
Scalar projection vector per axis is called number, the absolute value of which is equal to the length of the segment of the axis (in the selected scale) enclosed between the projections of the start point and the end point of the vector. Usually instead of the expression scalar projection simply say - projection. The projection is denoted by the same letter as the projected vector (in normal, non-bold writing), with a subscript (usually) of the name of the axis on which this vector is projected. For example, if a vector is projected onto the x-axis a, then its projection is denoted a x . When projecting the same vector onto another axis, if the axis is Y , its projection will be denoted as y .
To calculate projection vector on an axis (for example, the X axis) it is necessary to subtract the coordinate of the start point from the coordinate of its end point, that is
and x \u003d x k - x n.
The projection of a vector onto an axis is a number. Moreover, the projection can be positive if the value of x k is greater than the value of x n,
negative if the value of x k is less than the value of x n
and equal to zero if x k is equal to x n. 
The projection of a vector onto an axis can also be found by knowing the modulus of the vector and the angle it makes with that axis.
It can be seen from the figure that a x = a Cos α
that is, the projection of the vector onto the axis is equal to the product of the modulus of the vector and the cosine of the angle between the direction of the axis and vector direction. If the angle is acute, then
Cos α > 0 and a x > 0, and if obtuse, then the cosine obtuse angle is negative, and the projection of the vector onto the axis will also be negative.
Angles counted from the axis counterclockwise are considered to be positive, and in the direction - negative. However, since cosine is an even function, that is, Cos α = Cos (− α), when calculating projections, angles can be counted both clockwise and counterclockwise.
To find the projection of a vector onto an axis, the module of this vector must be multiplied by the cosine of the angle between the direction of the axis and the direction of the vector.
Vector coordinates are the coefficients of the only possible linear combination of basis vectors in the chosen coordinate system equal to the given vector.
The axis is the direction. Hence, the projection onto an axis or onto a directed line is considered the same. Projection can be algebraic or geometric. In geometric terms, the projection of a vector onto an axis is understood as a vector, and in algebraic terms, it is a number. That is, the concepts of the projection of a vector on an axis and the numerical projection of a vector on an axis are used.
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If we have an axis L and a non-zero vector A B → , then we can construct a vector A 1 B 1 ⇀ , denoting the projections of its points A 1 and B 1 .
A 1 B → 1 will be the projection of the vector A B → onto L .
Definition 1
The projection of the vector onto the axis a vector is called, the beginning and end of which are projections of the beginning and end of the given vector. n p L A B → → it is customary to denote the projection of A B → onto L . To construct a projection on L, drop the perpendiculars on L.
Example 1
An example of the projection of a vector onto an axis.
On the coordinate plane O x y, a point M 1 (x 1, y 1) is specified. It is necessary to build projections on O x and O y for the image of the radius vector of the point M 1 . Let's get the coordinates of the vectors (x 1 , 0) and (0 , y 1) .
If we are talking about the projection of a → onto a non-zero b → or the projection of a → onto the direction b → , then we mean the projection of a → onto the axis with which the direction b → coincides. The projection a → onto the line defined by b → is denoted n p b → a → → . It is known that when the angle is between a → and b → , we can consider n p b → a → → and b → codirectional. In the case when the angle is obtuse, n p b → a → → and b → are oppositely directed. In the situation of perpendicularity a → and b → , and a → is zero, the projection of a → along the direction b → is a zero vector.
The numerical characteristic of the projection of a vector onto an axis is the numerical projection of a vector onto a given axis.
Definition 2
Numerical projection of the vector onto the axis call a number that is equal to the product of the length of a given vector and the cosine of the angle between the given vector and the vector that determines the direction of the axis.
The numerical projection of A B → onto L is denoted n p L A B → , and a → onto b → - n p b → a → .
Based on the formula, we get n p b → a → = a → · cos a → , b → ^ , whence a → is the length of the vector a → , a ⇀ , b → ^ is the angle between the vectors a → and b → .
We get the formula for calculating the numerical projection: n p b → a → = a → · cos a → , b → ^ . It is applicable for known lengths a → and b → and the angle between them. The formula is applicable for known coordinates a → and b → , but there is a simplified version of it.
Example 2
Find out the numerical projection a → onto a straight line in the direction b → with the length a → equal to 8 and the angle between them is 60 degrees. By condition we have a ⇀ = 8 , a ⇀ , b → ^ = 60 ° . So, we substitute the numerical values into the formula n p b ⇀ a → = a → · cos a → , b → ^ = 8 · cos 60 ° = 8 · 1 2 = 4 .
Answer: 4.
With known cos (a → , b → ^) = a ⇀ , b → a → b → , we have a → , b → as scalar product a → and b → . Following from the formula n p b → a → = a → · cos a ⇀ , b → ^ , we can find the numerical projection a → directed along the vector b → and get n p b → a → = a → , b → b → . The formula is equivalent to the definition given at the beginning of the clause.
Definition 3
The numerical projection of the vector a → on the axis coinciding in direction with b → is the ratio of the scalar product of the vectors a → and b → to the length b → . The formula n p b → a → = a → , b → b → is applicable for finding the numerical projection of a → onto a straight line coinciding in direction with b → , with known a → and b → coordinates.
Example 3
Given b → = (- 3 , 4) . Find the numerical projection a → = (1 , 7) onto L .
Solution
On the coordinate plane n p b → a → = a → , b → b → has the form n p b → a → = a → , b → b → = a x b x + a y b y b x 2 + b y 2 , with a → = (a x , a y ) and b → = b x , b y . To find the numerical projection of the vector a → onto the L axis, you need: n p L a → = n p b → a → = a → , b → b → = a x b x + a y b y b x 2 + b y 2 = 1 (- 3) + 7 4 (- 3) 2 + 4 2 = 5 .
Answer: 5.
Example 4
Find the projection a → onto L , coinciding with the direction b → , where there are a → = - 2 , 3 , 1 and b → = (3 , - 2 , 6) . A three-dimensional space is given.
Solution
Given a → = a x , a y , a z and b → = b x , b y , b z calculate the scalar product: a ⇀ , b → = a x b x + a y b y + a z b z . We find the length b → by the formula b → = b x 2 + b y 2 + b z 2. It follows that the formula for determining the numerical projection a → will be: n p b → a ⇀ = a → , b → b → = a x b x + a y b y + a z b z b x 2 + b y 2 + b z 2 .
We substitute numerical values: n p L a → = n p b → a → = (- 2) 3 + 3 (- 2) + 1 6 3 2 + (- 2) 2 + 6 2 = - 6 49 = - 6 7 .
Answer: - 6 7 .
Let's look at the connection between a → on L and the length of the projection of a → on L . Draw an axis L by adding a → and b → from a point to L , after which we draw a perpendicular line from the end of a → to L and project onto L . There are 5 image variations:
The first the case when a → = n p b → a → → means a → = n p b → a → → , hence n p b → a → = a → cos (a , → b → ^) = a → cos 0 ° = a → = n p b → a → → .
Second case implies the use of n p b → a → ⇀ = a → cos a → , b → , so n p b → a → = a → cos (a → , b →) ^ = n p b → a → → .
Third case explains that when n p b → a → → = 0 → we get n p b ⇀ a → = a → cos (a → , b → ^) = a → cos 90 ° = 0, then n p b → a → → = 0 and n p b → a → = 0 = n p b → a → → .
Fourth case shows n p b → a → → = a → cos (180 ° - a → , b → ^) = - a → cos (a → , b → ^) , follows n p b → a → = a → cos (a → , b → ^) = - n p b → a → → .
Fifth case shows a → = n p b → a → → , which means a → = n p b → a → → , hence we have n p b → a → = a → cos a → , b → ^ = a → cos 180 ° = - a → = - n p b → a → .
Definition 4
The numerical projection of the vector a → on the axis L , which is directed like b → , has the meaning:
- the length of the projection of the vector a → onto L provided that the angle between a → and b → is less than 90 degrees or equal to 0: n p b → a → = n p b → a → → with the condition 0 ≤ (a → , b →) ^< 90 ° ;
- zero under the condition of perpendicularity a → and b → : n p b → a → = 0 when (a → , b → ^) = 90 ° ;
- the length of the projection a → onto L, times -1 when there is an obtuse or flattened angle of the vectors a → and b → : n p b → a → = - n p b → a → → with the 90° condition< a → , b → ^ ≤ 180 ° .
Example 5
Given the length of the projection a → onto L , equal to 2 . Find the numerical projection a → given that the angle is 5 π 6 radians.
Solution
It can be seen from the condition that this angle is obtuse: π 2< 5 π 6 < π . Тогда можем найти числовую проекцию a → на L: n p L a → = - n p L a → → = - 2 .
Answer: - 2.
Example 6
Given a plane O x y z with the length of the vector a → equal to 6 3 , b → (- 2 , 1 , 2) with an angle of 30 degrees. Find the coordinates of the projection a → onto the L axis.
Solution
First, we calculate the numerical projection of the vector a → : n p L a → = n p b → a → = a → cos (a → , b →) ^ = 6 3 cos 30 ° = 6 3 3 2 = 9 .
By condition, the angle is acute, then the numerical projection a → = is the length of the projection of the vector a → : n p L a → = n p L a → → = 9 . This case shows that the vectors n p L a → → and b → are co-directed, which means that there is a number t for which the equality is true: n p L a → → = t · b → . From here we see that n p L a → → = t b → , so we can find the value of the parameter t: t = n p L a → → b → = 9 (- 2) 2 + 1 2 + 2 2 = 9 9 = 3 .
Then n p L a → → = 3 b → with the coordinates of the projection of the vector a → onto the L axis are b → = (- 2 , 1 , 2) , where it is necessary to multiply the values by 3. We have n p L a → → = (- 6 , 3 , 6). Answer: (- 6 , 3 , 6) .
It is necessary to repeat the previously studied information about the condition of vector collinearity.
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Projecting various lines and surfaces onto a plane allows you to build a visual representation of objects in the form of a drawing. We will consider a rectangular projection, in which the projecting rays are perpendicular to the projection plane. PROJECTION OF A VECTOR ON A PLANE consider the vector \u003d (Fig. 3.22), enclosed between the perpendiculars dropped from its beginning and end.
 |
 |
Rice. 3.22. Vector projection of a vector onto a plane.
Rice. 3.23. Vector projection of a vector onto an axis.
In vector algebra, it is often necessary to project a vector onto an AXIS, that is, onto a straight line that has a certain orientation. Such a design is easy if the vector and axis L lie in the same plane (Fig. 3.23). However, the task becomes more difficult when this condition is not met. Let's construct the projection of the vector onto the axis, when the vector and the axis do not lie in the same plane (Fig. 3.24).
Rice. 3.24. Projecting a vector to an axis
in general.
Through the ends of the vector we draw planes perpendicular to the line L. At the intersection with this line, these planes define two points A1 and B1 - the vector, which we will call the vector projection of this vector. The problem of finding a vector projection can be solved more simply if the vector is brought into the same plane with the axis, which is possible, since free vectors are considered in vector algebra.
Along with the vector projection, there is also the SCALAR PROJECTION, which is equal to the module of the vector projection if the vector projection coincides with the orientation of the L axis, and is equal to the value opposite to it if the vector projection and the L axis have the opposite orientation. The scalar projection will be denoted by:
Vector and scalar projections are not always terminologically separated strictly in practice. The term "vector projection" is usually used, meaning by this the scalar projection of a vector. When deciding, it is necessary to clearly distinguish between these concepts. Following the established tradition, we will use the terms "vector projection", implying a scalar projection, and "vector projection" - in accordance with the established meaning.
Let us prove a theorem that allows us to calculate the scalar projection of a given vector.
THEOREM 5. The projection of a vector onto the axis L is equal to the product of its module and the cosine of the angle between the vector and the axis, that is
(3.5)
Rice. 3.25. Finding vector and scalar
Vector projections on the L axis
(and the L axis are equally oriented).
PROOF. Let us perform preliminary constructions that allow us to find the angle G Between the vector and the L axis. To do this, we construct a straight line MN parallel to the L axis and passing through the point O - the beginning of the vector (Fig. 3.25). The angle will be the desired angle. Let's draw through points A and O two planes perpendicular to the axis L. We get:
Since the L axis and the line MN are parallel.
We single out two cases of mutual arrangement of the vector and axis L.
1. Let the vector projection and the L axis be equally oriented (Fig. 3.25). Then the corresponding scalar projection 
.
2. Let and L be oriented in different sides(Fig. 3.26).
Rice. 3.26. Finding the vector and scalar projections of a vector on the L axis (and the L axis are oriented in opposite directions).
Thus, the assertion of the theorem holds in both cases.
THEOREM 6. If the beginning of the vector is reduced to a certain point of the L axis, and this axis is located in the s plane, the vector forms an angle with the vector projection onto the s plane, and an angle with the vector projection onto the L axis, in addition, the vector projections themselves form an angle between themselves , then
